## How do you take the complement of a regular language?

The complement of a DFA can be obtained by making the non-final states as final states and vice-versa. The language accepted by the complemented DFA L2 is the complement of the language L1. Lets first draw the DFA for L1 that accepts the strings of even length.

## Why is the complement of a regular language regular?

Remark 2: Since a language is regular if and only if it is accepted by some NFA, the complement of a regular language is also regular. Langauges are sets. Therefore all the properties of sets are inherited by languages. In particular De Morgan’s law also applies to languages.

**Is regular language closed under complement?**

The set of regular languages is closed under complementation. The complement of language L, written L, is all strings not in L but with the same alphabet.

**Why are regular languages closed under complement?**

Regular Languages are closed under complementation, i.e., if L is regular then L = Σ∗ \ L is also regular. Proof. If L is regular, then there is a DFA M = (Q,Σ, δ, q0,F) such that L = L(M). Then, M = (Q,Σ, δ, q0,Q \ F) (i.e., switch accept and non-accept states) accepts L.

### What is the complement of the language?

Defination: The complement of a language is defined in terms of set difference from Σ* (sigma star). that is L’ = Σ* – L. And the complement language (L’) of L has all strings from Σ* (sigma star) except the strings in L. Σ* is all possible strings over the alphabet Σ.

### Which language is accepted by finite automata?

Alternatively, a regular language can be defined as a language recognized by a finite automaton. The equivalence of regular expressions and finite automata is known as Kleene’s theorem (after American mathematician Stephen Cole Kleene).

**How do you prove a language is regular?**

To prove if a language is a regular language, one can simply provide the finite state machine that generates it. If the finite state machine for a given language is not obvious (and this might certainly be the case if a language is, in fact, non-regular), the pumping lemma for regular languages is a useful tool.

**Is the family of regular languages closed under infinite intersection?**

Each one is regular because it only contains one string. But the infinite union is the set {0i1i | i>=0} which we know is not regular. So the infinite union cannot be closed for regular languages.

#### What is meant by NFA?

NFA stands for non-deterministic finite automata. It is easy to construct an NFA than DFA for a given regular language. The finite automata are called NFA when there exist many paths for specific input from the current state to the next state. Every NFA is not DFA, but each NFA can be translated into DFA.

#### Can an infinite language be regular?

In the end, you can create infinite languages using finite descriptions (a regular expression). A finite language is a language containing a finite number of words. The simplest cases are those containing no words at all, the empty string, and a single string consisting of a single symbol (e.g. a in your example).

**Which type of language is accepted by PDA?**

The languages which can be accepted by PDA are called context-free languages (CFL), denoted by LCF. Diagrammatically, a PDA is a finite state automaton (see Fig. 5.1), with memories (push-down stacks).

**Is a * regular language?**

If a and b are regular expression, ab (concatenation of a and b) is also regular. If a is regular expression, a* (0 or more times a) is also regular. {a, ab, abb, abbb, abbbb,….} Regular Grammar : A grammar is regular if it has rules of form A -> a or A -> aB or A -> ɛ where ɛ is a special symbol called NULL.

## Why is the complement of a regular language still a?

Why is the complement of a regular language still a regular language? According to my textbook, the complement of L1 = A* – L1 is a regular language as long as L1 is a regular language. Doesn’t A* also include Context Free languages, Context Sensitive languages, and Recursively Enumerable languages?

## Which is a regular without complement without automata?

If you can show the intersection is a regular without complement and without automata that would be good as well. Alphabet Σ set of symbols, final. Word w finite concatenation of symbols from Σ and final. Sigma Kleene star Σ ∗ all words with symbols from Σ including ε the empty string. Language L subset of Σ ∗. Let L M be languages under Σ.

**Which is the most restricted type of language?**

As discussed in Chomsky Hierarchy, Regular Languages are the most restricted types of languages and are accepted by finite automata. Regular Expressions are used to denote regular languages.

**When are two regular expressions are the same?**

Note : Two regular expressions are equivalent if languages generated by them are same. For example, (a+b*)* and (a+b)* generate same language. For example, (a+b*)* and (a+b)* generate same language. Every string which is generated by (a+b*)* is also generated by (a+b)* and vice versa.